KVPYINTERVIEW )_(MATHEMATICS) SET - 1 1.
If all line segments are straight, in the given figure, then the sum of the angles at the corners marked in the diagram is : A
B
C
G F
D E
Sol.
2.
(A) 360º (B) 450º (C*) 540º (D) 630º Sum of the angles of the seven triangles = 180º × 7 = 1260º. The base angles of the triangle are the exterior angles of the seven-sided polygon. Now their sum = 2 × 360º = 720º the sum of the angles at the vertices marked = 1260º – 720º = 540º. In the adjoining figure, AB = AC, x = x1. The value of e in terms of a is : A a
x B
(A) e = 90º – a/2
(B*) e = 90º + a/2
E e
x1 C
(C) e = 180º – a
(D) e = 2a
Sol. A
x B
a
E e
c1
c
x1 C
As x = x1 (given) AECB is a cyclic quad. c = c1 c1 = 180º – e c1 =
(AE subtends eq. s at B and C) (base s, isos. ) (opp. s cyclic quad.)
180 º a 180 º a 180º – e = 2 2
180º – e = 90º –
a 2
a . 2 If X + Y + Z = 30, (X,Y,Z > 0), then the value of (X – 2) (Y– 3) (Z – 4) will be : (A) 1000 (B) 800 (C) 500 (D) 343 Let us consider the number X – 2, Y – 3, Z – 4. Their AM is GM. e = 90 +
3. Sol.
Therefore
[( X 2) ( Y 3) ( Z 4)] [(X – 2)(Y – 3)(Z – 4)]1/3 3
i.e.,[(X – 2)(Y – 3)(Z – 4)]1/3
XYZ9 3
PAGE # 1
i.e.,[(X – 2)(Y – 3)(Z – 4)]1/3
30 9 =7 3
On cubing both sides we get (X – 2)(Y – 3)(Z – 4) 73 (X – 2)(Y – 3)(Z – 4) 343. 4. Sol.
5.
The sum of the digits of the number (41000)(251002) is : (A) 7 (B) 8 (C) 11 41000 × 251002 22000 × 52004 54(2 × 5)2000 625(10)2000 625 × 1000.......(2000 times) = 625000.........(2000 times) Sum of digit = 6 + 2 + 5 = 13.
(D) 13
Line 2 intersects 1 and line 3 is parallel to 1 . The three lines are distinct and lie in a plane. The number of points equidistant from all the three lines is : (A) 0 (B) 1 (C*) 2 (D) 4
Sol.
Let draw line 4 which is parallel to 1 and 3 and equidistant from 1 and 3 The point which are equidistant from 1 and 1 is lie an the angle bisector, which intersect 4 at two points so there are two points which are eqidistant from all the three lines 1, 2 and 3. 6.
In the figure, AB is a diameter of the circle, TD is a tangent. If AHD = 36º, CDT is : D
T A
K
36º H
C
B
Sol.
(A) 100º (B) 110º (C) 116º ADB = 90º (in semi-circle) T DAH = 180º – 90º – 36º = 54º DAH = CDK (in alt. seg.) = 54º CDT = 180º – 54º (adj s on straight line) = 126º
(D) 126º D
K
36º H
A
C
B
7.
The minimum value of 2sinx + 2cos x is : (A) 1
Sol.
sinx
(B) 2
(C)
1 –
2
2
(D)
1
2
1 2
cos x
2 +2 As we know that sin x and cos x both lies between – 1 and + 1 but the values of sin x and cos x shows. Reverse pattern of each other so the sum will be minimum when both the values 2sin x and 2cos x are equal i.e. 2sin x = 2cos x i.e. sin x = cos x. i.e. x = /4. So the minimum value will be 2sin /4 + 2cos /4 = 2 ×
1
2
2
=
1
2
2
1
PAGE # 2
SET - 2 1.
In the figure, the quadrilateral ABCD is a rectangle, P lies on AD and Q and AB. The triangles PAQ, QBC and PCD all have the same area, and BQ = 2. The length of AQ, is : (A) 3 5
(B) 2 3
D
Sol.
(D) not uniquely determined
5 1
C
P a A
(C*)
x y
Q
B
2
ar BCQ = ar APQ
1 1 x × 2 = ay 2 2 2x = ay ay ... (i) 2 arBCQ = arPDC x=
1 1 ×2×x= (x – a)(y + 2) 2 2 2x = xy + 2x – ay – 2a xy = ay + 2a ay × y = ay + 2a 2
[Using (i)]
y2 =y+2 2 y2 = 2y + 4 y2 – 2y – 4 = 0
y=
2 4 16 2
2 20 22 5 = = 1 5 = 1 5 2 2 Ans. (C) because negative answer can’t be possible. =
2.
In the figure, AB = x. The area of triangle ADC is (angle B = 90º) D C A
30º 15º
B
1 2 1 1 x sin 30º (B) x2 tan 30º (C) x2 tan 45º 2 2 2 Area of triangle ADC = Area of triangle ABD – area of triangle ABC (A)
Sol.
x x = × BD – × BC 2 2 x = × (BD – BC) 2 x = x(x tan 45º – x tan 15º) 2 Area of triangle ADC =
(D)
1 2 x (tan 45º – tan 15º) 2 D C
A
30º 15º
B
1 2 ×x (tan 45º – tan 15º). 2 PAGE # 3
3. Sol.
In a single throw of 3 dice, the probability of not getting the same number on any 2 dice is (A) 0.45 (B) 0.66 (C) 0.56 (D) 0.83 (E) None of these P(not getting same number on 2 dice) = 1 – P(getting same number on 2 or 3 dice) sample space for getting same number on 2 or 3 dice = (1, 1, 1), (1, 1, 2), .....(1, 1, 6) (2, 2, 1), (2, 2, 2).....(2, 2, 6) . . . (6, 6, 1), (6, 6, 2)......(6, 6, 6) So total 36 cases. P (not getting same number on any two dice)
=1–
36 216
1 5 = = 0.83 6 6 Two parallel lines are one unit apart. A circle of radius 2 touches one of the lines and cuts the other line.
=1–
4.
The area of the circular cap between the two parallel lines can be written in from of
Sol.
(a + b) of the two integers a and b equal to : (A) 3 (B) 4 In triangle AOD
(C*) 5
(D) 6 C
1 cos = 2 = 60º AOB = 2 = 120º
D1 1
A 2
B 2
O
area of segment ACB = =
a – b 3 . The sum 3
2 1 r2 – r2sin2 360 º 2
120 º 2 1 2 r – r sin120 360 º 2
4 a – 3 = –b 3 3 3 a=4 b=1 a + b = 4 + 1 = 5.
5.
In the figure, CD is the diameter of a semicircle CBED with centre O, and AB = OD. If EOD = 60º, then BAC is E B A
(A) 15º
C
(B*) 20º
O
(C) 30º
D
(D) 45º
E B
Sol. A
x
60º
x C
O
D
AB = OD = OE = OB = OC (radius) BOC = BAC = x (AB = OB) OBE = x + x =2x (Exterion angle of ABO) OEB = OBE = 2x (OB = OE) EOD = x + 2x = 3x = 60º x = 20º.
PAGE # 4
6.
Sol.
7.
Let , be the roots of the equation (x – a)(x – b) = c, c 0. Then the roots of the equation (x – )(x – ) + c = 0 are : (A) a, c (B) b, c (C*) a, b (D) a + c, b + c (x – a)(x – b) = c x2 – x(a + b) + ab – c = 0 + =a + b = ab – c ab = + c x2 – x(+ ) + + c = 0 x2 – x(+ ) + ab = 0 So, the roots are a and b. Option (C) is the answer. On a plane are two points A and B at a distance of 5 unit apart. The number of straight lines in this plane which are at distance of 2 units from A and 3 units from B, is : (A) 1 (B) 2 (C*) 3 (D) 4 L1
L3
Sol.
B
A
L2
Three straight lines L1, L2 & L3 can be drawn as shown in figure, which are at the distance 2 unit from A and 3 unit from B.
SET - 3 1.
Circles with centres O, O’ and P each tangent of the line L and also mutually tangent. If the radii of circle O and circle O’ are equal and the radius of the circle P is 6, then the radius of the larger circle is :
O'
O P
(A) 22
(B) 23
L
(C*) 24
(D) 25
Sol. R
O A
R
O' R–6 R 6 L
R
R P
OP = R + 6, OA = R – 6 and AP = R. OP2 = OA2 + AP2 (R + 6)2 = (R – 6)2 + R2 R2 + 12R + 36 = R2 – 12R + 36 + R2 24 R = R2 So, R = 24 unit. 2.
In the diagram B, C and D lie on a straight line, with ACD = 100°, ADB = x°, ABD = 2x° and DAC = BAC = y°. The value of (sin y°. tan y° + sec y°) equals :
A y° y° 2x°
B (A*) 7/2
100°
C (B) 3
x° D (C) 5/2
(D) 5
PAGE # 5
Sol.
80 = x + y 100 = 2x + y – – – – 20 = – x x = 20 y = 60 sin y. tan y + sec y = sin 60 × tan 60 + sec 60 = =
3.
3 × 2
y° y° 2x° 80° 100°
B
C
x° D
3 +2
3 7 +2= . 2 2
1 1 1 + + loga bc 1 logb ca 1 logc ab 1 = (A) 1
Sol.
A
(B) 2
(C) 3
(D) 4
1 1 1 + + loga bc loga a logb ca logb b logc ab logc c 1 1 1 = log abc + log abc + log abc a b c = logabc a + logabc b + logabc c = logabc abc = 1.
4.
Sol.
5.
The sides of a triangle with positive area have lengths 4, 6 and x. The sides of a second triangle with positive area have length 4, 6 and y. The smallest positive number that is not the possible value of |x – y|, is (x and y are integers) (A) 2 (B) 4 (C) 6 (D*) 8 As we know sum of two side in always greater than third side 4+6>x 10 > x 4+6>y 10 > y And we also know that the difference of two side in always less than the third side. 6–4
Sol.
(3a 2b) 18c
(B)
( 2a b ) 24c
(C)
(a 4b) 6c
(D)
(a b ) 3c
Part of container C filled with water amount of water in amount of water in amount of water in + container B + container C container A
=
3c
a b 0 3a 2b 2 3 = = . 3c 18c
PAGE # 6
6. Sol.
The number of common terms of the two sequences 17, 21, 25, ....., 417 and 16, 21, 26, .... 466 is : (A) 21 (B) 19 (C) 20 (D) 91 17, 21, 25, 29, 33, 37, 41, ....., 417 16, 21, 26, 31, 36, 41, ...., 466 We know that tn = a + (n – 1)d Now the common terms are 21, 41, 61, ....., 401 Set tn = 401 401 = 21 + (n – 1)20 401 – 21 = (n – 1)20 380 =n–1 20 19 = n – 1 20 = n.
380 = (n – 1) 20
7.
Each of the congruent circles shown is externally tangent to other circles and/or to the side(s) of the rectangle as shown. If each circle has circumference 16, then the length of a diagonal of the rectangle, is
(A*) 80
(B) 40
(C) 20
(D) 15
Sol.
6r r 8r diagonal =
(8r )2 (6r )2 = 10r
2r = 16 r =8 Diagonal = 10 × 8 = 80.
SET - 4 1. Sol.
2.
The GCD of two numbers is 13 and their product is 4732. The possible number of pairs is/are (A) 0 (B) 1 (B) 2 (D) 3 Let the numbers are 13P and 13Q. Product is PQ × 132 = 4732. PQ = 28. As P and Q are co-prime P = 1, Q = 28 or P = 4, Q = 7. In the triangle ABC, AB = 6, BC = 5, CA = 4. AP bisects the angle A and P lies on BC. Then AP equals (A) 3.75
Sol.
(B) 3.1
(C) 2.9
(D) 4.2
As AP is angle bisector of BAC BP = 3, PC = 2 In ABP,
sin sin = 3 x
In ABC,
sin 2 sin = 5 4
2 sin cos 24 x sin = x= cos 5 5 12
A
6
4 x
B
3
P 5
2
C
PAGE # 7
36 16 25 27 62 42 52 = = 48 48 2 6 4
cos 2 =
cos2 =
75 1 cos 2 = 96 2
24 7.07 × = 4.2 x= 5 8
3.
Three parallel lines 1, 2 and 3 are drawn through the vertices A, B and C of a square ABCD. If the distance between 1 and 2 is 7 and between 2 and 3 is 12, then the area of the square ABCD is : (A*) 193 (B) 169 (C) 196 (D) 225 C D
12
Sol.
7
H 12
7
A
G
Let a be the side of square : In triangle CHB 12 a In triangle ABG
cos =
7 a 2 2 sin + cos = 1
cos(90 – ) =
144 a 4.
2
+
49 a2
7 a
a2 = 193
=1
Given x = 1 + a + a2 + ......... and y = 1 + b + b2 + ...... where a and b are proper functions. 1 + ab + a2b2 + ..... equal to
xy (A) x y 1 Sol.
sin =
(B)
x2y2 xy
xy (C) x y
xy
(D)
2
x y2
Here x is sum of an infinite series with first term as 1 and common ratio a 1 x 1 x(1 – a) = 1 x – xa = 1 a = 1 a x Similarly
x=
y=
1 y (1 – b) = 1 y – yb = 1 y – 1 = yb 1 b
b =
y 1 x 1 y . Now ab = x
y 1 xy x y 1 = y xy
Now, s = 1+ ab + a2b2 + ......
1 xy xy 1 s= = 1 xy x y 1 = xy xy x y 1 = x y 1 1 ab xy
PAGE # 8
5.
The probability that a teacher will given a surprise test during a class meeting is
1 . If a student is absent 5
on two days, then what is the probability that he will miss at least one test ? 9 16 (B) 25 25 P(at least one test) = 1 – P(no test)
(A) Sol.
(C)
4 5
(D)
2 5
1 1 16 9 ) × (1 – ) = 1 – = 5 5 25 25 day 1 day 2
= 1 – (1 –
6.
In an army during a war, 4 men out of every 25, were wounded and 2 out of every 25 were killed. 38000 men returned unhurt. What was the original number of men in the army ? (A) 50000
Sol.
(B) 37500
(C) 28200
(D) 56550
Let the total men in army = x. Out of every 25, 4 are wounded. Number of men wounded out of x =
4x 25
2 men were killed out of every 25. Number of men killed out of x =
2x 25
Total number of men = Number of men wounded + number of men killed + unhurt men x=
4x 2x + + 38000 25 25
19x = 950000 x = 50000. 7.
If x > y > 0 and 2 log(x – y) = logx + logy, then x/y equals : (A) 3 +
Sol.
5
(B*)
3 5 only 2
(C)
3 5 only 2
(D)
3 5 2
2 log (x–y) = logx + logy (x–y)2 = xy x2 + y2 – 3xy = 0 2
3x x – +1=0 y y
(divide by y2)
x 3 9–4 3 5 = = y 2 2 since x > y
x >1 y
x 3 5 = y 2
PAGE # 9
SET - 5 1.
A, B, C and D walk towards a point P taking the shortest path from the four vertices of a rectangle. After they reach the point P, which is inside the rectangle, it is found that the distance travelled by A is 20m, B is 30m, C is 25 m (as shown in the figure). Find the distance travelled by D. A
D 20 P
25
30 B
(A) 40m Sol.
(B) 25m 2
(C) 15 3
2
2
It can be obtained that a + x = 20 b2 + x2 = 302 b2 + d2 = 252 eq.(i) – eq.(ii) + eq.(iii) a2 + d2 = 202 – 302 + 252 But a2 + d2 = y2 (from DEP) y = DP =
2.
C
...(i) ...(ii) ...(iii)
20 2 30 2 25 2 = 5 5 .
A
(D) None of these x E 20 a P
b
b 30 B x F
d
D
y 25 d
C
If x x x x ....... 2 then x equals : (A*) 2 – 2
(B) 2 2 x 2 = 2
(C) 2 2
(D) 2 – 3
Sol.
x x x x ....... 2
x = 2 – 2 .
3.
All the 3 sides of a right triangle are integers and one side has a length 11 units. Area of the triangle in square units lies between : (A) 1 and 100 (B) 100 and 200 (C) 200 and 300 (D*) More than 300
Sol.
c a
b = 11 2
c = a2 + b2 c2 – a2 = 112 c2 – a2 = 121 (c + a) (c – a) = 121 × 1 c + a = 121 c–a=1 2c = 122 c = 61 a = 60
Area =
1 1 ab = × 60 × 11 = 330. 2 2
PAGE # 10
4.
ABCD is a parallelogram, M is the midpoint of DC. If AP = 65 and PM = 30 then the largest possible integral value of AB is : (A*) 124 A B (B) 120 (C) 119 (D) 118 P
Sol.
APB ~ CPM
D
M
A
C
65
AP PB AB = = CP PM CM 65 PB 2CM = = CP 30 CM PB = 60 In APB AB < 60 + 65 AB < 125 AB = 124 is the largest integral value. 5.
Sol.
6. Sol.
B P 30
D
M
C
Six friends are sitting around a campfire. Each person in turn announces the total of the ages of the other five people. If 104, 105, 108, 114, 115 and 119 given the six sums of each group of five people. The age of the oldest person (A*) is 29 (B) is 30 (C) is 31 (D) can not be found out Let the six friend are x1, x2, x3, x4, x5, x6 x2 + x3 + x4 + x5 + x6 = 104 ...(i) x1 + x3 + x4 + x5 + x6 = 105 ...(ii) x1 + x2+ x4 + x5 + x6 = 108 ...(iii) x1 + x2 + x3 + x5 + x6 = 114 ...(iv) x1 + x2 + x3 + x4 + x6 = 115 ...(v) x1 + x2 + x3 + x4 + x5 = 119 ...(vi) Add all 5(x1 + x2 +x3 + x4 + x5 + x6) = 665 x1 + x2 + x3 + x4 + x5 + x6 = 133 – (vii) to get maximum age person subtract (i) from (vii) We get x1 = 29. Let f(x) = 2x. For which value of x is f(x–2) = f(x) – 2 ? (A) 8/3 (B) 1.415 (C) 2 2x – 2, = 2x – 2. Let y = 2x – 2 y = 4y – 2 3y = 2 y = 2/3 y = 2x – 2 = 2/3 x = 2 + log2(2/3) = 2 + log22 – log23 = 3 – log2 3.
(D) 3 – log2 3
3 3 2 – 2 when simplified reduces to 2 2 (A) an irrational number (B) a whole number (C*) a rational which is not an integer (D) a natural number
7.
The expression 2
Sol.
2
3 3 2 – 2 2 2 2
3 1 – 2 2 1 2 2
PAGE # 11
1 – 3 – 21 2 2 2+ =
3 2 – 2 –
2
2
1 So it is a rational. 2
SET - 6 1.
Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is m times the area of the square. The ratio of the area of the other small right triangle to the area of the square is :
1 4m arEFC = m(a2) (A*)
Sol.
(B)
1 2m 1
(C) m
(D)
1 8m 2
A
1 × a × x = ma2 2 x = 2am ADE ~ ABC
y D
y a = ya a 2am
a
ya y =1+2m
E
a a
B
a
F
x
C
a 1+y =1+2m y=
a 2m
1 a a 2m = 1 . Req. ratio 2 2 4m a
2.
A and B can demolish a building in 3 days. B and C can do it in 6 days. C and A in 5 days. If A, B and C work together and C gets injured at the end of first day work and can not come back, then the total number of days to complete the work by A and B will be : (A)
Sol.
60 13
(B*)
59 20
(C)
Work done by A and B in one day =
1 part 3
Work done by B and C in one day =
1 part 6
Work done by C and A in one day =
1 part 5
Work done by A, B & C in 1 days =
1 2
=
20 7
(D) 3
1 1 1 3 6 5
1 10 5 6 30 2 PAGE # 12
=
21 7 1 × = part 30 20 2 7 part 20
Work done by A, B and C in 1 day =
7 13 = part. 20 20 Now, A & B can do complete work together in 3 days.
Remaining work = 1 –
So, they can do
13 part in 20
=3×
13 39 = days 20 20
39 59 = days. 20 20 In parallelogram ABCD, the length AB and CD are both 4, the length of diagonal AC = 4, and the length of diagonal BD = 6. The length AD equal to :
Total number of days to complete the work = 1 + 3.
(A*) Sol.
10 arABC = ar ABD
(B)
(C) 15
12
(D)
4
D
x x x x 5 1 1 5 = 2 2 2 2 2 2 2 25 x x 1 = x 4 4 4
x x x x 4 4 ` 2 2 2 2
4.
C
x
4
A
2 16 x 4
2 100 x 2 x 2 4 2 = x 64 x 4 4 4 4 100x2 – 400 – x4 + 4x2 = 64x2 – x4 40x2 = 400 x2 = 10
x=
x
20
6 4
B
10 .
In the figure C is a right angle, DE AB, A E = 6, EB = 7 and BC = 5. The area of the quadrilateral EBCD is C D
A
(A) 27.5 Sol.
(B) 25
6
E
5 B
7
(C*) 22.5
(D) 20 C
AC =
13 2 5 2 = 12 AED ~ ACB (By AA)
AE ED = AC CB
D
A
6
E
5
7
B
ED 6 = 5 12 ED = 2.5 ar of quadrilateral EBCD = area ABC – area AED PAGE # 13
5.
1 1 × 5 × 12 – × 6 × 2.5 2 2 = 30 – 7.5 = 22.5 In the right triangle shown the sum of the distances BM and MA is equal to the distances BC and CA. If MB = x, CB = h and CA = d, then x equals. M B
A
C
hd (B) d – h (C) h + d – 2d 2h d ATQ. x + y = h + d M y=h+d–x 2 2 2 y = (x + h) + d x (h + d – x)2 = (x + h)2 + d2 B y h2 + d2 + x2 + 2hd – 2dx – 2hx = x2 + h2 + 2xh + d2 h 2hd – 2dx – 2xh = 2xh hd = xh + dx + xh C d hd = x(2h + d)
(A*) Sol.
x= 6.
h 2 d2 h
A
hd . 2h d
Given triangle PQR with RS bisecting R, PQ extended to D and n a right angle, then
(A) m =
1 (p – q) 2
(B) m =
1 (p + q) 2
(C) d =
1 (q + p) 2
R
m
P
1 m 2 x + m = n = 90º In PSR RSQ = x + p In RSQ x + x + p + q = 180º 2x + p + q = 180º 2(90º – m) + p + q = 180º 180º – 2m + p + q = 180º 2m = p + q
n q
p
d
D
Q
S
(D) d = Sol.
(D)
R x x m
P
n = 90 q
p S
d
D
Q
pq . 2 The sides of rectangle are all produced in order, in such a way that the length of each side is increased m=
7.
by ‘k’ times itself. The area of the new quadrilateral formed becomes 2
1 times the area of the original 2
rectangle. The value of ‘k’ is : (A)
1 3
(B)
1 4
(C*)
1 2
(D)
2 3
PAGE # 14
IV
Sol.
D A
x
ky II C y III B kx
The new figure obtained is made up of the original rectangle and four additional triangles, I, II, III, IV, as marked in the figure in which, I = II and III = IV. If x and y are the sides of the original rectangle, the sides of triangle I or II are x(1 + k), ky and of triangle III or IV are y(1 + k), kx. Areas of I + II + III + IV
= 2[
1 1 x (1 + k) ky + y(1 + k)kx] 2 2
= k(1 + k)xy + k(1 + k)xy = 2k(1 + k)xy. The area of the new quadrilateral = 2k(1 + k)xy + xy =
So, 2k(1 + k) + 1 =
5 xy (given). 2
5 3 giving k(k + 1) = , 2 4
i.e. 4k2 + 4k – 3 = 0,(2k + 3)(2k – 1) = 0 i.e., k =
1 3 1 or – which is impossible. Hence the required k = . 2 2 2
PAGE # 15
